prepare 20 ml of 10% soln using: 1st 2%, 2nd 0.5%, 3rd 11%, 4th 6%, 5th 5%, 6th 3%

[Nirmeen]

Dear friends i need help
I have 6 water-based 10% solutions
and i want to prepare 20 ml using the following percentage:

the first 2%
the second 0.5%
the third 11%
the fourth 6%
the fifth 5%
the sixth 3%

also i want to prepare 50 ml using the following percentage:

the first 12%
the second 0.5 %
the third 3%
the fourth 1%
the fifth 5%
the sixth 3%

could somebody tell me how to do that
thanks in advance and your help would be appreciated

 

[stapel]

I have 6 water-based 10% solutions

Then aren't these solutions all the same?

and i want to prepare 20 ml using the following percentage:

the first 2%
the second 0.5%
the third 11%
the fourth 6%
the fifth 5%
the sixth 3%

You want to prepare 20 mL of what by using the following percentage of what?

also i want to prepare 50 ml using the following percentage:

the first 12%
the second 0.5 %
the third 3%
the fourth 1%
the fifth 5%
the sixth 3%

Please reply with the full and exact text of the exercise, the complete instructions, and a clear listing of your thoughts and efforts so far. Also, since you posted this to "Arithmetic", clearly you haven't studied any pre-algebra or algebra, so we can't assume that you know anything about variables or systems of equations. So, when you reply, please include information on whatever grammar-school method you've been given for this sort of exercise.

Thank you! :wink:

 

[Nirmeen]

Thanks stapel for your reply and please excuse my poor English
I have 6 different types of solutions
all of them are water-based
and all of them are 10 percent solutions " like saline 10% solution for example"

and now i need to mix up 20 ml of those 6 different types of solutions in one bottle using the following percentage
of each one of those 6 solutions
2% of the first solution
0.5% of the second solution
11% of the third solution
6% of the fourth solution
5% of the fifth solution
3% of the sixth solution
so how much ml should i use of each one of these 6 different solutions to mix up 20 ml.

Also i need to mix up 50 ml of those 6 different types of solutions in one bottle using the following percentage
of each one of those 6 solutions
12% of the first solution
0.5% of the second solution
3% of the third solution
1% of the fourth solution
5% of the fifth solution
3% of the sixth solution
so how much ml should i use of each one of these 6 different solutions to mix up 50 ml.

thanks again wishing you a good day

 

[stapel]

I have 6 different types of solutions
all of them are water-based
and all of them are 10 percent solutions " like saline 10% solution for example"

and now i need to mix up 20 ml of those 6 different types of solutions in one bottle using the following percentage
of each one of those 6 solutions
2% of the first solution
0.5% of the second solution
11% of the third solution
6% of the fourth solution
5% of the fifth solution
3% of the sixth solution
so how much ml should i use of each one of these 6 different solutions to mix up 20 ml.

Are you saying that, of the 20 mL of mixture, 2% of it should come from Solution 1's bottle? If so, then note that 2% + 0.5% + 11% + 6% + 5% + 3% does not equal 100%, so you would not be able to obtain 20 mL of anything, using the specified inputs.

Or are you saying that you need to use 2% of whatever is in the bottle of Solution 1? So, if Solution 1 has 100 mL, you should use 2 mL of Solution 1? If so, we need to know how many milliliters are in each bottle.

Either way, this exercise makes very little sense.

please excuse my poor English

Please have somebody better versed in the language provide a corrected translation of the original exercise, so we can try to figure out what is going on here. Thank you!

 

[HallsofIvy]

Thanks stapel for your reply and please excuse my poor English
I have 6 different types of solutions
all of them are water-based
and all of them are 10 percent solutions " like saline 10% solution for example"

So one is 10% sodium chloride, one is 10% potassium chlorate, another is 10% sodium hydroxide, etc?

and now i need to mix up 20 ml of those 6 different types of solutions in one bottle using the following percentage
of each one of those 6 solutions
2% of the first solution
0.5% of the second solution
11% of the third solution
6% of the fourth solution
5% of the fifth solution
3% of the sixth solution
so how much ml should i use of each one of these 6 different solutions to mix up 20 ml.

if you let X be the amount, in ml, of each of these you are saying that you want
0.02X+ 0.005X+ .11X+ .06X+ .05X+ .03X= 0.215X= 20. So X= 20/0.215= 93 ml.

Also i need to mix up 50 ml of those 6 different types of solutions in one bottle using the following percentage
of each one of those 6 solutions
12% of the first solution
0.5% of the second solution
3% of the third solution
1% of the fourth solution
5% of the fifth solution
3% of the sixth solution
so how much ml should i use of each one of these 6 different solutions to mix up 50 ml.

thanks again wishing you a good day

So now you have 0.12X+ .005X+ 0.03X+ 0.01X+ 0.05X+ 0.03X= 0.245X= 50 so X= 50/0.245= 204 ml.

Those are assuming you take the same amount, X ml, of each solution. Otherwise there are an infinite number of possible solutions.